1

Chapter 4

GreedyAlgorithms

Slides by Kevin Wayne.Copyright © 2005 Pearson-Addison Wesley.All rights reserved.

4.1 Interval Scheduling

3

Interval Scheduling

Interval scheduling. Job j starts at sj and finishes at fj. Two jobs compatible if they don't overlap. Goal: find maximum subset of mutually compatible jobs.

Time0 1 2 3 4 5 6 7 8 9 10 11

f

g

h

e

a

b

c

d

4

Interval Scheduling: Greedy Algorithms

Greedy template. Consider jobs in some order. Take each job provided it's compatible with the ones already taken.

[Earliest start time] Consider jobs in ascending order of start time sj.

[Earliest finish time] Consider jobs in ascending order of finish time fj.

[Shortest interval] Consider jobs in ascending order of interval length fj - sj.

[Fewest conflicts] For each job, count the number of conflicting jobs cj. Schedule in ascending order of conflicts cj.

5

Interval Scheduling: Greedy Algorithms

Greedy template. Consider jobs in some order. Take each job provided it's compatible with the ones already taken.

breaks earliest start time

breaks shortest interval

breaks fewest conflicts

6

Greedy algorithm. Consider jobs in increasing order of finish time. Take each job provided it's compatible with the ones already taken.

Implementation. O(n log n). Remember job j* that was added last to A. Job j is compatible with A if sj fj*.

Sort jobs by finish times so that f1 f2 ... fn.

A for j = 1 to n { if (job j compatible with A) A A {j}}return A

jobs selected

Interval Scheduling: Greedy Algorithm

7

Interval Scheduling: Analysis

Theorem. Greedy algorithm is optimal.

Pf. (by contradiction) Assume greedy is not optimal, and let's see what happens. Let i1, i2, ... ik denote set of jobs selected by greedy. Let j1, j2, ... jm denote set of jobs in the optimal solution with

i1 = j1, i2 = j2, ..., ir = jr for the largest possible value of r.

j1 j2 jr

i1 i1 ir ir+1

. . .

Greedy:

OPT: jr+1

why not replace job jr+1

with job ir+1?

job ir+1 finishes before jr+1

8

j1 j2 jr

i1 i1 ir ir+1

Interval Scheduling: Analysis

Theorem. Greedy algorithm is optimal.

Pf. (by contradiction) Assume greedy is not optimal, and let's see what happens. Let i1, i2, ... ik denote set of jobs selected by greedy. Let j1, j2, ... jm denote set of jobs in the optimal solution with

i1 = j1, i2 = j2, ..., ir = jr for the largest possible value of r.

. . .

Greedy:

OPT:

solution still feasible and optimal, but contradicts maximality of r.

ir+1

job ir+1 finishes before jr+1

4.1 Interval Partitioning

10

Interval Partitioning

Interval partitioning. Lecture j starts at sj and finishes at fj. Goal: find minimum number of classrooms to schedule all

lectures so that no two occur at the same time in the same room.

Ex: This schedule uses 4 classrooms to schedule 10 lectures.

Time9 9:30 10 10:30 11 11:30 12 12:30 1 1:30 2 2:30

h

c

b

a

e

d g

f i

j

3 3:30 4 4:30

11

Interval Partitioning

Interval partitioning. Lecture j starts at sj and finishes at fj. Goal: find minimum number of classrooms to schedule all

lectures so that no two occur at the same time in the same room.

Ex: This schedule uses only 3.

Time9 9:30 10 10:30 11 11:30 12 12:30 1 1:30 2 2:30

h

c

a e

f

g i

j

3 3:30 4 4:30

d

b

12

Interval Partitioning: Lower Bound on Optimal Solution

Def. The depth of a set of open intervals is the maximum number that contain any given time.

Key observation. Number of classrooms needed depth.

Ex: Depth of schedule below = 3 schedule below is optimal.

Q. Does there always exist a schedule equal to depth of intervals?

Time9 9:30 10 10:30 11 11:30 12 12:30 1 1:30 2 2:30

h

c

a e

f

g i

j

3 3:30 4 4:30

d

b

a, b, c all contain 9:30

13

Interval Partitioning: Greedy Algorithm

Greedy algorithm. Consider lectures in increasing order of start time: assign lecture to any compatible classroom.

Implementation. O(n log n). For each classroom k, maintain the finish time of the last job

added. Keep the classrooms in a priority queue.

Sort intervals by starting time so that s1 s2 ... sn.d 0

for j = 1 to n { if (lecture j is compatible with some classroom k) schedule lecture j in classroom k else allocate a new classroom d + 1 schedule lecture j in classroom d + 1 d d + 1 }

number of allocated classrooms

14

Interval Partitioning: Greedy Analysis

Observation. Greedy algorithm never schedules two incompatible lectures in the same classroom.

Theorem. Greedy algorithm is optimal.Pf.

Let d = number of classrooms that the greedy algorithm allocates.

Classroom d is opened because we needed to schedule a job, say j, that is incompatible with all d-1 other classrooms.

Since we sorted by start time, all these incompatibilities are caused by lectures that start no later than sj.

Thus, we have d lectures overlapping at time sj + . Key observation all schedules use d classrooms. ▪

4.2 Scheduling to Minimize Lateness

16

Scheduling to Minimizing Lateness

Minimizing lateness problem. Single resource processes one job at a time. Job j requires tj units of processing time and is due at time dj. If j starts at time sj, it finishes at time fj = sj + tj. Lateness: j = max { 0, fj - dj }. Goal: schedule all jobs to minimize maximum lateness L =

max j.

Ex:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

d5 = 14d2 = 8 d6 = 15 d1 = 6 d4 = 9d3 = 9

lateness = 0lateness = 2

dj 6

tj 3

1

8

2

2

9

1

3

9

4

4

14

3

5

15

2

6

max lateness = 6

17

Minimizing Lateness: Greedy Algorithms

Greedy template. Consider jobs in some order.

[Shortest processing time first] Consider jobs in ascending order of processing time tj.

[Earliest deadline first] Consider jobs in ascending order of deadline dj.

[Smallest slack] Consider jobs in ascending order of slack dj - tj.

18

Greedy template. Consider jobs in some order.

[Shortest processing time first] Consider jobs in ascending order of processing time tj.

[Smallest slack] Consider jobs in ascending order of slack dj - tj.

counterexample

counterexample

dj

tj

100

1

1

10

10

2

dj

tj

2

1

1

10

10

2

Minimizing Lateness: Greedy Algorithms

19

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

d5 = 14d2 = 8 d6 = 15d1 = 6 d4 = 9d3 = 9

max lateness = 1

Sort n jobs by deadline so that d1 d2 … dn

t 0for j = 1 to n Assign job j to interval [t, t + tj] sj t, fj t + tj

t t + tj

output intervals [sj, fj]

Minimizing Lateness: Greedy Algorithm

Greedy algorithm. Earliest deadline first.

20

Minimizing Lateness: No Idle Time

Observation. There exists an optimal schedule with no idle time.

Observation. The greedy schedule has no idle time.

0 1 2 3 4 5 6

d = 4 d = 6

7 8 9 10 11

d = 12

0 1 2 3 4 5 6

d = 4 d = 6

7 8 9 10 11

d = 12

21

Minimizing Lateness: Inversions

Def. An inversion in schedule S is a pair of jobs i and j such that:i < j but j scheduled before i.

Observation. Greedy schedule has no inversions.

Observation. If a schedule (with no idle time) has an inversion, it has one with a pair of inverted jobs scheduled consecutively.

ijbefore swap

inversion

22

Minimizing Lateness: Inversions

Def. An inversion in schedule S is a pair of jobs i and j such that:i < j but j scheduled before i.

Claim. Swapping two adjacent, inverted jobs reduces the number of inversions by one and does not increase the max lateness.

Pf. Let be the lateness before the swap, and let ' be it afterwards.

'k = k for all k i, j 'i i If job j is late:

ij

i j

before swap

after swap

f'j

fi

inversion

23

Minimizing Lateness: Analysis of Greedy Algorithm

Theorem. Greedy schedule S is optimal.Pf. Define S* to be an optimal schedule that has the fewest number of inversions, and let's see what happens.

Can assume S* has no idle time. If S* has no inversions, then S = S*. If S* has an inversion, let i-j be an adjacent inversion.

– swapping i and j does not increase the maximum lateness and strictly decreases the number of inversions

– this contradicts definition of S* ▪

24

Greedy Analysis Strategies

Greedy algorithm stays ahead. Show that after each step of the greedy algorithm, its solution is at least as good as any other algorithm's.

Exchange argument. Gradually transform any solution to the one found by the greedy algorithm without hurting its quality.

Structural. Discover a simple "structural" bound asserting that every possible solution must have a certain value. Then show that your algorithm always achieves this bound.